3.683 \(\int \frac{\sqrt{-1+x^2} \sec ^{-1}(x)}{x^2} \, dx\)

Optimal. Leaf size=107 \[ \frac{i \sqrt{x^2} \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(x)}\right )}{x}-\frac{i \sqrt{x^2} \text{PolyLog}\left (2,i e^{i \sec ^{-1}(x)}\right )}{x}-\frac{1}{\sqrt{x^2}}-\frac{\sqrt{x^2-1} \sec ^{-1}(x)}{x}-\frac{2 i \sqrt{x^2} \sec ^{-1}(x) \tan ^{-1}\left (e^{i \sec ^{-1}(x)}\right )}{x} \]

[Out]

-(1/Sqrt[x^2]) - (Sqrt[-1 + x^2]*ArcSec[x])/x - ((2*I)*Sqrt[x^2]*ArcSec[x]*ArcTan[E^(I*ArcSec[x])])/x + (I*Sqr
t[x^2]*PolyLog[2, (-I)*E^(I*ArcSec[x])])/x - (I*Sqrt[x^2]*PolyLog[2, I*E^(I*ArcSec[x])])/x

________________________________________________________________________________________

Rubi [A]  time = 0.153779, antiderivative size = 116, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {5242, 4698, 4710, 4181, 2279, 2391, 8} \[ \frac{i \sqrt{x^2} \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(x)}\right )}{x}-\frac{i \sqrt{x^2} \text{PolyLog}\left (2,i e^{i \sec ^{-1}(x)}\right )}{x}-\frac{1}{\sqrt{x^2}}-\frac{\sqrt{1-\frac{1}{x^2}} \sqrt{x^2} \sec ^{-1}(x)}{x}-\frac{2 i \sqrt{x^2} \sec ^{-1}(x) \tan ^{-1}\left (e^{i \sec ^{-1}(x)}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-1 + x^2]*ArcSec[x])/x^2,x]

[Out]

-(1/Sqrt[x^2]) - (Sqrt[1 - x^(-2)]*Sqrt[x^2]*ArcSec[x])/x - ((2*I)*Sqrt[x^2]*ArcSec[x]*ArcTan[E^(I*ArcSec[x])]
)/x + (I*Sqrt[x^2]*PolyLog[2, (-I)*E^(I*ArcSec[x])])/x - (I*Sqrt[x^2]*PolyLog[2, I*E^(I*ArcSec[x])])/x

Rule 5242

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Dist[Sqrt[x
^2]/x, Subst[Int[((e + d*x^2)^p*(a + b*ArcCos[x/c])^n)/x^(m + 2*(p + 1)), x], x, 1/x], x] /; FreeQ[{a, b, c, d
, e, n}, x] && IGtQ[n, 0] && EqQ[c^2*d + e, 0] && IntegerQ[m] && IntegerQ[p + 1/2] && GtQ[e, 0] && LtQ[d, 0]

Rule 4698

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((
f*x)^(m + 1)*Sqrt[d + e*x^2]*(a + b*ArcCos[c*x])^n)/(f*(m + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1 -
c^2*x^2]), Int[((f*x)^m*(a + b*ArcCos[c*x])^n)/Sqrt[1 - c^2*x^2], x], x] + Dist[(b*c*n*Sqrt[d + e*x^2])/(f*(m
+ 2)*Sqrt[1 - c^2*x^2]), Int[(f*x)^(m + 1)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}
, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 1])

Rule 4710

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Dist[(c^(m +
1)*Sqrt[d])^(-1), Subst[Int[(a + b*x)^n*Cos[x]^m, x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ
[c^2*d + e, 0] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{-1+x^2} \sec ^{-1}(x)}{x^2} \, dx &=-\frac{\sqrt{x^2} \operatorname{Subst}\left (\int \frac{\sqrt{1-x^2} \cos ^{-1}(x)}{x} \, dx,x,\frac{1}{x}\right )}{x}\\ &=-\frac{\sqrt{1-\frac{1}{x^2}} \sqrt{x^2} \sec ^{-1}(x)}{x}-\frac{\sqrt{x^2} \operatorname{Subst}\left (\int 1 \, dx,x,\frac{1}{x}\right )}{x}-\frac{\sqrt{x^2} \operatorname{Subst}\left (\int \frac{\cos ^{-1}(x)}{x \sqrt{1-x^2}} \, dx,x,\frac{1}{x}\right )}{x}\\ &=-\frac{1}{\sqrt{x^2}}-\frac{\sqrt{1-\frac{1}{x^2}} \sqrt{x^2} \sec ^{-1}(x)}{x}+\frac{\sqrt{x^2} \operatorname{Subst}\left (\int x \sec (x) \, dx,x,\sec ^{-1}(x)\right )}{x}\\ &=-\frac{1}{\sqrt{x^2}}-\frac{\sqrt{1-\frac{1}{x^2}} \sqrt{x^2} \sec ^{-1}(x)}{x}-\frac{2 i \sqrt{x^2} \sec ^{-1}(x) \tan ^{-1}\left (e^{i \sec ^{-1}(x)}\right )}{x}-\frac{\sqrt{x^2} \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(x)\right )}{x}+\frac{\sqrt{x^2} \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(x)\right )}{x}\\ &=-\frac{1}{\sqrt{x^2}}-\frac{\sqrt{1-\frac{1}{x^2}} \sqrt{x^2} \sec ^{-1}(x)}{x}-\frac{2 i \sqrt{x^2} \sec ^{-1}(x) \tan ^{-1}\left (e^{i \sec ^{-1}(x)}\right )}{x}+\frac{\left (i \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sec ^{-1}(x)}\right )}{x}-\frac{\left (i \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \sec ^{-1}(x)}\right )}{x}\\ &=-\frac{1}{\sqrt{x^2}}-\frac{\sqrt{1-\frac{1}{x^2}} \sqrt{x^2} \sec ^{-1}(x)}{x}-\frac{2 i \sqrt{x^2} \sec ^{-1}(x) \tan ^{-1}\left (e^{i \sec ^{-1}(x)}\right )}{x}+\frac{i \sqrt{x^2} \text{Li}_2\left (-i e^{i \sec ^{-1}(x)}\right )}{x}-\frac{i \sqrt{x^2} \text{Li}_2\left (i e^{i \sec ^{-1}(x)}\right )}{x}\\ \end{align*}

Mathematica [A]  time = 0.172135, size = 116, normalized size = 1.08 \[ -\frac{\sqrt{1-\frac{1}{x^2}} \left (-i x \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(x)}\right )+i x \text{PolyLog}\left (2,i e^{i \sec ^{-1}(x)}\right )+\sqrt{1-\frac{1}{x^2}} x \sec ^{-1}(x)-x \sec ^{-1}(x) \log \left (1-i e^{i \sec ^{-1}(x)}\right )+x \sec ^{-1}(x) \log \left (1+i e^{i \sec ^{-1}(x)}\right )+1\right )}{\sqrt{x^2-1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sqrt[-1 + x^2]*ArcSec[x])/x^2,x]

[Out]

-((Sqrt[1 - x^(-2)]*(1 + Sqrt[1 - x^(-2)]*x*ArcSec[x] - x*ArcSec[x]*Log[1 - I*E^(I*ArcSec[x])] + x*ArcSec[x]*L
og[1 + I*E^(I*ArcSec[x])] - I*x*PolyLog[2, (-I)*E^(I*ArcSec[x])] + I*x*PolyLog[2, I*E^(I*ArcSec[x])]))/Sqrt[-1
+ x^2])

________________________________________________________________________________________

Maple [B]  time = 0.391, size = 708, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(x)*(x^2-1)^(1/2)/x^2,x)

[Out]

1/4*(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*x^3-3*I*x^2-4*((x^2-1)/x^2)^(1/2)*x+4*I)/(-I*((x^2-1)/x^2)^(1/2)*x+x^2-
1)/x-1/2*(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*x-I)*x/(-I*((x^2-1)/x^2)^(1/2)*x+x^2-1)-1/4/(x^2-1)^(1/2)/x*(I*((x
^2-1)/x^2)^(1/2)*x^3-2*I*((x^2-1)/x^2)^(1/2)*x+2*x^2-2)*arcsec(x)+1/4/(x^2-1)^(1/2)*(I*((x^2-1)/x^2)^(1/2)*x^3
-2*I*((x^2-1)/x^2)^(1/2)*x-2*x^2+2)*arcsec(x)/x+I*(x^2-1)^(1/2)*x^3/(4*I*((x^2-1)/x^2)^(1/2)*x^3-8*I*((x^2-1)/
x^2)^(1/2)*x+8*x^2-8)-1/2/(x^2-1)^(1/2)*(I*x^2+((x^2-1)/x^2)^(1/2)*x-I)*arcsec(x)*ln(1+I*(1/x+I*(1-1/x^2)^(1/2
)))+1/2/(x^2-1)^(1/2)*(I*x^2+((x^2-1)/x^2)^(1/2)*x-I)*arcsec(x)*ln(1-I*(1/x+I*(1-1/x^2)^(1/2)))-1/2/(x^2-1)^(1
/2)*(-I*((x^2-1)/x^2)^(1/2)*x+x^2-1)*dilog(1+I*(1/x+I*(1-1/x^2)^(1/2)))+1/2/(x^2-1)^(1/2)*(-I*((x^2-1)/x^2)^(1
/2)*x+x^2-1)*dilog(1-I*(1/x+I*(1-1/x^2)^(1/2)))+1/2/(x^2-1)^(1/2)*(I*x^2-((x^2-1)/x^2)^(1/2)*x-I)*arcsec(x)*ln
(1+I*(1/x+I*(1-1/x^2)^(1/2)))-1/2/(x^2-1)^(1/2)*(I*x^2-((x^2-1)/x^2)^(1/2)*x-I)*arcsec(x)*ln(1-I*(1/x+I*(1-1/x
^2)^(1/2)))+1/2/(x^2-1)^(1/2)*(I*((x^2-1)/x^2)^(1/2)*x+x^2-1)*dilog(1+I*(1/x+I*(1-1/x^2)^(1/2)))-1/2/(x^2-1)^(
1/2)*(I*((x^2-1)/x^2)^(1/2)*x+x^2-1)*dilog(1-I*(1/x+I*(1-1/x^2)^(1/2)))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} - 1} \operatorname{arcsec}\left (x\right )}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)*(x^2-1)^(1/2)/x^2,x, algorithm="maxima")

[Out]

integrate(sqrt(x^2 - 1)*arcsec(x)/x^2, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{2} - 1} \operatorname{arcsec}\left (x\right )}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)*(x^2-1)^(1/2)/x^2,x, algorithm="fricas")

[Out]

integral(sqrt(x^2 - 1)*arcsec(x)/x^2, x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\left (x - 1\right ) \left (x + 1\right )} \operatorname{asec}{\left (x \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(x)*(x**2-1)**(1/2)/x**2,x)

[Out]

Integral(sqrt((x - 1)*(x + 1))*asec(x)/x**2, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} - 1} \operatorname{arcsec}\left (x\right )}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)*(x^2-1)^(1/2)/x^2,x, algorithm="giac")

[Out]

integrate(sqrt(x^2 - 1)*arcsec(x)/x^2, x)